Chosen maximal subject is a subgroup

Question

Let $ G $ is a finite soluble group and $ N $ be a unique minimal normal subgroup of $ G $. Let $ G = TS $ that $ S $ is the fitting subgroup of $ G $ and $ T = N_{G}(H) $ for $ H \leq G $. Suppose $ T \cap N = 1 $. Let $ U $ be chosen maximal subject to $ T \leq U $ and $ U \cap N = 1 $. Why $ U < G $ ?

Answer 1

It seems that what you're saying is that $U$ is supposed to be a subgroup of $G$ that is maximal with respect to the following property:

$T\le U$ and $U\cap N=1.$

If we had $U=G,$ then we would have $U\cap N=N\ne 1,$ so clearly we can't have $U=G.$ Moreover, since (it seems) you're requiring that $U\leq G,$ then we have $U<G.$

Observe that this has nothing to do with the properties of $G,$ nor with $S,T,$ or $H.$ We don't even need $N$ to be unique! It seems that we can rephrase as follows:

If $G$ is any group, $N$ any minimal normal subgroup of $G,$ $T$ any subgroup of $G$ such that $T\cap N=1,$ and $U$ is a subgroup of $G$ that is maximal with respect to the conditions $T\subseteq U$ and $U\cap N=1,$ then $U<G.$